There has been a number of helpful threads clarifying the essence of the notion of the opposite category (particularly here and here). However I am still confused with taking the opposite of the category where sets of morphisms between two objectes are themselves objects of the category (e.g. Set or R-mod).

Here is the source of confusion:

Let us take the category $\mathcal{C}$ to be R-mod for commutative $R$ with identity.

Now for $R$-modules $M,N$, the set $\operatorname{Hom}_{R-mod}(M,N)$ is again an $R$-module and for any morphism $M_1 \xrightarrow{\phi} M_2$ we have the induced maps $$(*) \ \ \ \ \ \ \operatorname{Hom}_{R-mod} (M_1, N) \xleftarrow{\phi^*} \operatorname{Hom}_{R-mod} (M_2, N),$$ $$\ \ \ \ \ \ \ \ \ \ \ \operatorname{Hom}_{R-mod} (P, M_1) \xrightarrow{\phi_*} \operatorname{Hom}_{R-mod} (P, M_2),$$ which are $R$-linear, so that they are morphisms in $\mathcal{C}$ = R-mod.

Therefore in $\mathcal{C}^{op}$ = R-mod$^{op}$ we obtain a morphisms "opposite" to $\phi^*$ and $\phi_*$:

$$(**) \ \ \ \ \ \ \operatorname{Hom}_{R-mod^{op}}(N^{op}, M_{1}^{op}) \xrightarrow{({\phi^{*}})^{op}}\operatorname{Hom}_{R-mod^{op}}(N^{op}, M_{2}^{op}),$$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname{Hom}_{R-mod^{op}}(M_{1}^{op}, P^{op}) \xleftarrow{({\phi_*})^{op}}\operatorname{Hom}_{R-mod^{op}}(M_{2}^{op}, P^{op}),$$

where I am using the identification $(\operatorname{Hom}_{R-mod}(X, Y))^{op} = \operatorname{Hom}_{Y^{op}}(Y^{op}, X^{op})$.

On the other hand, the opposite to $\phi$, $M_1^{op} \xleftarrow{\phi^{op}} M_2^{op}$, induces maps in the inverse directions to $(**)$:

$$(***) \ \ \ \ \ \ \operatorname{Hom}_{R-mod^{op}}(N^{op}, M_{1}^{op}) \xleftarrow{{(\phi^{op}})_*}\operatorname{Hom}_{R-mod^{op}}(N^{op}, M_{2}^{op}),$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \operatorname{Hom}_{R-mod^{op}}(M_{1}^{op}, P^{op}) \xrightarrow{{(\phi^{op}})^*}\operatorname{Hom}_{R-mod^{op}}(M_{2}^{op}, P^{op}),$$

  1. Do I understand it correctly, that $(***)$ are just set-maps and not morphisms in R-mod$^{op}$, since there is no reason to believe that there are corresponding morphisms $$\operatorname{Hom}_{R-mod} (M_1, N) \to \operatorname{Hom}_{R-mod} (M_2, N)$$ $$\operatorname{Hom}_{R-mod} (P,M_1) \leftarrow \operatorname{Hom}_{R-mod} (P, M_2)$$ in R-mod (these directions are unnatural)?
  2. If 1 is true, how then to make sense that in R-mod morphisms between objects induce morphisms between Hom's (see $(*)$), but in R-mod$^{op}$ we merely get set-theoretic maps $(***)$ which are not morphisms in $\mathcal{C}^{op}$? Doesn't that seem unnatural, since in my mind the opposite category should mimic the properties of $\mathcal{C}$ very closely.
  3. Closely related to 2, it seems that R-mod comes equipped with just a pair of maps $(*)$, whereas R-mod$^{op}$ comes equipped with two pairs, $(**)$ and $(***)$, which also seems strange.

PS. My goal is to understand the fact that contravariant $h_N=\operatorname{Hom}_{R-mod} ( - , N)$ is right-adjoint to itself meaning that we view $h_N :$ R-mod $\to $ R-mod$^{op}$ as right adjoint to $h_N^{op}:$ R-mod$^{op}$ $\to$ R-mod as explained in the algebra textbook of Paolo Aluffi. On the level that we "just have to reverse all arrows to account for contravariant nature of $h_N$" I understand his argument, but rigorous understanding of what is going on in the opposite categories evades me.


The proper way to understand this is not by fixing a single $\phi$. Let $C$ be the category of $R$-modules and $C^*$ its opposite. You have a composition map of modules, $C(N,P)\otimes C(M,N)\to C(M,P), f\otimes g\mapsto f\circ g$. In the opposite category, we have $C^*(P,N)=C(N,P)$ (where we don't distinguish between a module viewed in $C$ versus $C^*$.) So the above composition becomes a map (in $C$!) $C^*(P,N)\otimes C^*(N,M)\to C^*(P,M)$. Applying the symmetry isomorphism for the tensor product of modules, this is essentially the same as a map (still in $C$!) $C^*(N,M)\otimes C^*(P,N)\to C^*(P,M)$, which is just the composition map for $C^*$. Every hom-object and map thereof I've mentioned is an $R$-module, so we see $C^*$ still has homs and composition maps from $C$.

It's also perfectly possible to view $\otimes$ as a functor $C^*\times C^*\to C^*$. But that doesn't mean that the same composition map makes $C^*$'s composition be a map of $C^*$-objects: it goes in the wrong direction for that! So in (**), it wasn't meaningful to make the morphisms of hom-modules lie in $C^*$: you should leave them in $C$. This is not so unfamiliar: after all, the opposite of a category is a category, not something with homs from complete atomic Boolean algebras (that is, from the opposite of sets!)

In (***), you're missing something entirely. No such morphisms exist, as you almost prove right away: there are neither module maps nor set functions $C(M_1,N)\to C(M_2,N)$ induced by $\phi$. What could they possibly be?

The situation we're summarizing is leading toward the general theory of "enriched categories", which are splice categories but with homs forming something more interesting than a set. It's always the case that, when defined, the opposite of an enriched category is enriched over the same category, not a new one.

Anyway, the adjunction you're trying to describe has nothing to do with any of these issues. Let $F(M)=C(M,N)$ as an $R$-module, Then $C(P,F(M))=C(M,F(P))$, which by definition is $C^*(F(P),M)$. There's really no need for deep reflection on opposite categories here, although it might well be worthwhile in its own right.

  • $\begingroup$ thank you for your answer. I don't understand your comment about (***): isn't it true that for any category a given $M_1 \xrightarrow{\phi} M_2$ induces set-theoretic maps on sets of morphisms (which form sets by axioms of a category) by pulling-back (precomposing - $\alpha \mapsto \alpha \circ \phi$) or post-composing ($\beta \mapsto \phi \circ \beta$)? How is $\mathcal{C}^{op}$ different in this regard? $\endgroup$ – Bananeen Nov 14 '16 at 4:12
  • $\begingroup$ Oh, I misread what you wrote. The arrows you drew are just maps in $C$, the same ones I described. The thing you draw in question 1 is a map of $C^*$: you're flipping too much. $\endgroup$ – Kevin Carlson Nov 14 '16 at 17:40

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